I was sick a while back and I woke up in the middle of the night thinking about the “Stacking Bricks” problem. The problem asks how far out can you extend if you stack bricks like this.

The answer is: As far as you want. I will show this later. If you allow tricky stacking like this

you get all sorts of cool patterns. You also get farther out for a given number of bricks. See this paper by Paterson, Peres, Thorup, Winkler, and Zwick.

So I had two “brilliant” ideas. First I thought that you could get the span to be longer just by turning each brick so that a corner stuck out not an edge. It didn’t take me too long to figure out that the stacking brick problem is really a two dimensional problem – that the bricks are really tiles.

#### My next thought and the content of the rest of this post is what would happen if instead of stacking the bricks exactly balanced at their center of gravity, which is practically impossible, each time we inserted another brick on the bottom we give ourselves an to be safe. This can be seen in this diagram.

My guess was that the “safety factor” would be enough to limit the growth of the overhang to a finite number. I was wrong.

First the ideal case. This image shows my derivation of the fact that the nth inserted brick increases the span by .

Note that my “proof” is by implicit mathematical induction. This means that the span can be made as long as you want, since the sum of a harmonic series is infinite. This was proved as long ago as c. 1360 by Nicole Oresme.

This image shows my exploration of using a “safety” factor, .

It turns out that instead of we get . The span grows without bound but more slowly.

I think this problem could be extended to inhomogeneous bricks balanced at their center of gravities.